JEE Advanced | 2013ThermochemistryMedium

Question

The standard enthalpies of formation of CO2(g), H2O(l) and glucose(s) at 250C are -400 kJ/mol, -300 kJ/mol & -1300 kJ/mol respectively. The standard enthalpy of combustion per gram of glucose at 250C is -

Options

A.+ 2900 kJ
B.- 2900 kJ
C.- 16.11 kJ
D.+ 16.11 kJ

Solution

Combustion reaction for glucose is
C6H12O6(s) + 6O2(g)→ 6CO2(g) + 6H2O(l)
ᐃHr = 6(-400) + 6(-300) - (-1300)
= -2900 KJ/mol KJ / gm
= -16.11 KJ/gm

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